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PROFESSOR: Welcome back
to recitation.
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In this video, what I'd like
us to do is answer the
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following question.
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Suppose that f is a continuous,
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differentiable function.
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And if it's derivative, if f
prime is never 0, and a is not
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equal to b, then show that f of
a is not equal to f of b.
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I'm going to let you think about
it for a while, see if
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you can come up with a good
reason for that, and then I'll
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be back to explain my reasons.
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OK.
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Our object, again, is to show,
if f is a continuous and
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differentiable and its
derivative is never 0 and
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you're looking at two x values
that are different, show that
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their y values have
to be different.
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Show that if the inputs are
different, the outputs have to
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be different.
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Now, this might remind you of
something you saw in lecture
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about if the derivative has a
sign, show the function, if
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the derivative is positive, show
the function is always
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increasing, or if the derivative
is negative, show
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the function is always
decreasing.
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So this is a similar type
of problem to that.
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So what we're going
to use is actually
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the mean value theorem.
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If you'll notice, I have f.
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It does satisfy the mean
value theorem on an
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interval from a to b.
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I haven't even specified which
is bigger, a or b.
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But it doesn't matter
in this case.
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So what do we know?
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The mean value theorem tells us
that if we look at-- well,
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let's just write it out-- f of
b minus f of a over b minus a
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is equal to f prime of c--
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and what do we know-- for
c between a and b.
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So we want to know whether
or not f of b minus f of
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a can ever be 0.
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We're trying to show that
it cannot be 0.
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So we're going to isolate this
expression and show that this
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subtraction cannot be 0.
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Well, how do we do that?
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Let me come over here to give
us a little more room.
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I'm going to rewrite the
mean value theorem.
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I'm going to multiply through
by b minus a.
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So we get f prime of
c time b minus a.
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Now, we just want to show,
again, that f of b minus f of
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a cannot be 0.
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What's the only thing-- well,
not only thing, we know two
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things-- what two things
do we know?
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We know f prime of c is not 0.
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That was given to you.
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f prime is never 0, so certainly
at any fixed value,
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f prime of c is not 0.
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So we know this term is not 0.
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We also know that b is not equal
to a, so we know b minus
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a is not 0.
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The only way to get a product
of two numbers to be 0 is if
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one of them is 0.
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So this in fact, this product
is not equal to 0.
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The fact that this product is
not equal to 0 tells us f of b
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minus f of a is not
equal to 0.
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And that alone is enough to
conclude that f of b is not
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equal to f of a.
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So, again, let me just point
out of this is probably
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reminds you very much of the
type of thing you've seen
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where you were showing if f
prime had a sign, then you
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could determine whether f was
increasing or decreasing.
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It's the same type of
problem as that.
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It's exploiting what the mean
value theorem tells you.
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So I think we'll stop there.
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